Let $P$ be the point on line segment $\overline{AB}$ such that $AP:PB = 3:2.$  Then
\[\overrightarrow{P} = t \overrightarrow{A} + u \overrightarrow{B}\]for some constants $t$ and $u.$  Enter the ordered pair $(t,u).$

[asy]
unitsize(1 cm);

pair A, B, P;

A = (0,0);
B = (5,1);
P = interp(A,B,3/5);

draw(A--B);

dot("$A$", A, S);
dot("$B$", B, S);
dot("$P$", P, S);
[/asy]
Solution: Since $AP:PB = 3:2,$ we can write
\[\frac{\overrightarrow{P} - \overrightarrow{A}}{3} = \frac{\overrightarrow{B} - \overrightarrow{P}}{2}.\]Isolating $\overrightarrow{P},$ we find
\[\overrightarrow{P} = \frac{2}{5} \overrightarrow{A} + \frac{3}{5} \overrightarrow{B}.\]Thus, $(t,u) = \boxed{\left( \frac{2}{5}, \frac{3}{5} \right)}.$